485. Max Consecutive Ones

https://leetcode.com/problems/max-consecutive-ones/

easy 題，思路不說了，直接上代碼。

class Solution {public int findMaxConsecutiveOnes(int[] nums) {int cnt = 0;int max = 0;for (int num : nums) {if (num == 0) cnt = 0;else max = Math.max(max, ++cnt);}return max;} }

487. Max Consecutive Ones II

力扣這題需要 plus 會員，所以，為了保證“Max Consecutive Ones”的完整性，臨時用一下 lintcode。

https://www.lintcode.com/problem/883/

本題思路由題目 III 加了一個條件 k=1 改編而來。

另外，lintcode 的測試用例不夠全面，之前寫了一個明顯的 bug 但還是通過了，后來才發現，已更正。

public class Solution {public int findMaxConsecutiveOnes(int[] nums) {int k = 1;int cnt = 0;List<Integer> list = new ArrayList<>();for (int n : nums) {if (n == 0) {if (cnt != 0) list.add(cnt);list.add(0);cnt = 0;} else cnt++;}if (nums[nums.length - 1] == 1) list.add(cnt);int maxSum = 0;for (int i = 0; i < list.size(); i++) {int curSum = 0;int zeroCnt = 0;for (int j = i; j < list.size(); j++) {if (list.get(j) == 0) {if (++zeroCnt > k) break;curSum++;} else {curSum += list.get(j);}}maxSum = Math.max(maxSum, curSum);}if (maxSum == 0) {int zeroCnt = 0;for (Integer n : list) {if (n == 0) {if (++zeroCnt > k) break;maxSum++;} else {maxSum += n;}}}return maxSum;} }

1004. Max Consecutive Ones III

https://leetcode.com/problems/max-consecutive-ones-iii/

普通思路

public class Solution {public int longestOnes(int[] nums, int k) {// 建立輔助數組int cnt = 0;List<Integer> list = new ArrayList<>();for (int n : nums) {if (n == 0) {if (cnt != 0) list.add(cnt);list.add(0);cnt = 0;} else cnt++;}if (nums[nums.length - 1] == 1) list.add(cnt); // 處理邊界// 利用輔助數組計算maxSumint maxSum = 0;for (int i = 0; i < list.size(); i++) {int curSum = 0;int zeroCnt = 0;for (int j = i; j < list.size(); j++) {if (list.get(j) == 0) {if (++zeroCnt > k) break;curSum++;} else {curSum += list.get(j);}}maxSum = Math.max(maxSum, curSum);}// 獨立處理因數組長度不足而無法進入for循環的情況if (maxSum == 0) {int zeroCnt = 0;for (Integer n : list) {if (n == 0) {if (++zeroCnt > k) break;maxSum++;} else {maxSum += n;}}}return maxSum;} }

進階思路：滑動窗口

思路參考：https://leetcode-cn.com/problems/max-consecutive-ones-iii/solution/hua-dong-chuang-kou-de-liang-chong-jie-j-8ses/

雙指針法，時間復雜度O(n)。

public int longestOnes(int[] A, int K) {int left = 0;//窗口左邊的位置int right = 0;//窗口右邊的位置int zeroCount = 0;//窗口中0的個數for (; right < A.length; right++) {if (A[right] == 0) {zeroCount++;}//如果窗口中0的個數超過了K，要縮小窗口的大小if (zeroCount > K && A[left++] == 0)zeroCount--;}return right - left; }

總結

以上是生活随笔為你收集整理的leetcode 485,487,1004. Max Consecutive Ones I ,II, III（最大连续1的个数问题合集）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。